The polynomial on the right-hand side has a double zero at the origin and a zero at \(x = 2\).

Further, \[\begin{equation*} \frac{dy}{dx} = -3x^2 + 4x = x(4-3x) \end{equation*}\]so the curve has two stationary points, at \(x = 0\) and \(x = \dfrac{4}{3}\).

Since the leading term of the polynomial is \(-x^3\), \(y \to \infty\) as \(x \to -\infty\), and \(y \to -\infty\) as \(x \to \infty\).

Consequently, the first stationary point must be a local minimum and the second stationary point must be a local maximum.

This information, when put together, leads to the following sketch.

indicating *briefly* how the form of the curve has been derived.

Firstly, note that this curve does not exist where the right hand side is negative, that is when \(x>2\). This corresponds to where the curve in the first part of the question is below the \(x\)-axis.

For other values of \(x\), the curve has two branches, \(y=\pm x\sqrt{2-x}\).

The curve cuts or touches the \(x\)-axis at \(x=0\) and at \(x=2\).

So we have some kind of loop above and below the \(x\)-axis between \(x=0\) and \(x=2\), while for \(x<0\) the curve is in two branches, one increasing and the other decreasing.

In order to discover the nature of the axis crossings, we could think about how \(y=\sqrt{2-x}\) behaves and what happens when we multiply it by \(x\). Alternatively, we can do some calculus.

Differentiating implicitly we find \[\begin{align*} y^2 &=2x^2-x^3 \\ \implies\quad 2y\frac{dy}{dx} &= 4x-3x^2 \\ \iff\quad \frac{dy}{dx} &=\frac{4x-3x^2}{2y} \end{align*}\]As \(x\) approaches \(2\), \(y\to 0\) and \(\frac{dy}{dx}\to\frac{8-12}{2y}\to\pm\infty\). So a tangent to the curve at \(x=2\) would be vertical.

We can’t evaluate \(\frac{dy}{dx}\) at the origin, but nearby \(x\) is very small so \(y^2\approx2x^2\). This means that \(y\approx\pm\sqrt{2}x\) and we have \[\frac{dy}{dx}\approx\frac{4x-3x^2}{\pm 2\sqrt{2}x} \approx \pm\sqrt{2}.\] Hence the curve passes through the origin with gradient \(\pm\sqrt{2}\).

We are led, then, to the following sketch.